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@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I got it by one method
But lets say if i want to generalize it
Multinomial
Thm. or beggars method
Wth is beggards method
Jee mein nahi ha aisa kuch
That (n-r-1)C(r-1) thing
Oh thats beggar
Are you have read about it you just don't know the name
No i didnt know the name
(Shoutout to Opt who told me this lol)
Lets say agar n side ka.polygon ha
But coming back, think about the number of points between 2 points.
Now uska ek point uthane ke tareeke n
The sum of all of these would be n-k if n sided polygon has to be marked with k points non-consecutive
Ab uske saath vale nahi uthane to uske tareeke n-3
So you have a_1+a_2+..,a(k)=n-k
And then since all these are greater than equal to 1
Apply formula with b(i)+1=a(i)
Ab teesre vale ke liye problem ha
@SirLancelotDuLac jese mene likha ha usse 3rd point ka pata kese chlega
Nikal toh sakte hain the answer will be a summation
But its much more convenient by beggar's
I'll check this and get back to it tommorrow
Im revising pnc, to beggars abhi aane vala ha, agar usme dikkat ayi to ping krdunga
basically equivalent to asking how many triangles (doesnt matter if congruent or not)(even congruent ones are distinct) can be made by selecting 3 vertices such that no 2 vertices form the side of the polygon
popular q
so what u can do is
(no. of ways of choosing 3 vertices) - (no. of sides * no. of vertices left after subtracting 2)
now first part should be quite clear
(no. of sides * no. of vertices left after subtracting 2)this might be a bit confusing but let me clear it by choosing a side u guarantee choosing 2 consecutive vertices (which we want to subtract from the total) now wth is this
no. of vertices left after subtracting 2now for 3rd vertice u cant choose from the same 2 vertices (its expected that u choose distinct vertices) so ur pool of selection is no. of vertices - 2 (which u already used in a side) now this will give u all the possible triangles having 2 or more consecutive vertices simply subtract and ur done derived it for general polygon 10C3 - (10*8) 120 - 80 40
Choose any of the n points, which can be done in n tareeke.
Then let this be point $a_1$, the next one be $a_2$ and so on till $a_k$. The number of points between various $a_i's$ add up till (n-k), so call the distance between $ai$ and $a(i+1)$ as $b_i$ and the distance between $a_n$ and $a_1$ as $b_n$.
Now you have $b_1+b_2+...b_n=(n-k)$. Write $b_i=1+c_i$ and apply multinomial on $c_1+c_2+c_3+...c_k=(n-2k)$, which gives $\binom{(n-k-1)}{(k-1)}$. But now, each case is repeated k times, so the final generalization for this is $\frac{n \cdot \binom{(n-k-1)}{(k-1)}}{k}$
SirLancelotDuLac
K baar repition isliye kyunki you can think about how choosing a different point as a_1 would affect this
@Prasan
Okk
Smjh gyaa
Damn nicr method
+solved @SirLancelotDuLac
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